1. Understanding the Problem:
We need to prove that for any prime $p > 11$, there exists an integer $k$ such that the product $p \cdot k$ can be written in the decimal system using only the digit 1. This means we need to find $k$ such that $p \cdot k$ is a repunit number (a number consisting only of the digit 1).

2. Properties of Repunit Numbers:
A repunit number in decimal form can be written as:
$$R_n = \frac{10^n - 1}{9}$$
where $n$ is the number of digits (all being 1).

3. **Finding $k$:**
We need to find $k$ such that:
$$p \cdot k = R_n = \frac{10^n - 1}{9}$$
for some integer $n$.

4. Using Modular Arithmetic:
Since $p$ is a prime and $p > 11$, we know that $\gcd(10, p) = 1$. By Fermat's Little Theorem, we have:
$$10^{p-1} \equiv 1 \pmod{p}$$
This implies:
$$10^{p-1} - 1 \equiv 0 \pmod{p}$$
Therefore, $10^{p-1} - 1$ is divisible by $p$.

5. Constructing the Repunit:
Consider the number:
$$R_{p-1} = \frac{10^{p-1} - 1}{9}$$
Since $10^{p-1} - 1$ is divisible by $p$, we can write:
$$10^{p-1} - 1 = p \cdot m$$
for some integer $m$. Thus:
$$R_{p-1} = \frac{p \cdot m}{9}$$

6. **Finding $k$:**
We need $k$ such that:
$$p \cdot k = R_{p-1}$$
From the above, we have:
$$k = \frac{R_{p-1}}{p} = \frac{\frac{10^{p-1} - 1}{9}}{p} = \frac{10^{p-1} - 1}{9p}$$

7. Conclusion:
Therefore, the number $k = \frac{10^{p-1} - 1}{9p}$ satisfies the condition that $p \cdot k$ is a repunit number.

$$\boxed{k = \frac{10^{p-1} - 1}{9p}}$$